Problem: The grades on a geometry midterm at Almond are normally distributed with $\mu = 85$ and $\sigma = 3.5$. Stephanie earned a $96$ on the exam. Find the z-score for Stephanie's exam grade. Round to two decimal places.
Answer: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Stephanie's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{96 - {85}}{{3.5}}} $ ${ z \approx 3.14}$ The z-score is $3.14$. In other words, Stephanie's score was $3.14$ standard deviations above the mean.